All categories

3 years ago

I really need help please!

1 answer:

5 0

Simple...

you have: $\frac{ 3^{-4}* 2^{3}* 3^{2} }{ 2^{4}* 3^{-3} }$

Multiply it out-->>

$\frac{8}{9}$ ÷ $\frac{16}{27}$

= $\frac{3}{2}$ (A)

Thus, your answer.

You might be interested in

$400 interest is earned on a principal of $2,000 at a simple interest rate of 5% interest per year. For how many years was the p

bogdanovich [222]

Answer:

4 years

Step-by-step explanation:

Given data

Simple interest= $400

Principal= $2000

Rate= 5%

The simple interest formula is given as

SI=PRT/100

substitute and solve for T

400=2000*5*T/100

cross multiply

400*100= 10000*T

40000=10000T

T= 40000/10000

T= 4 years

Hence the time taken is 4 years

4 0

2 years ago

Which numbers are 4 units from -2 on the number line

Artemon [7]

Let the number be x

Then the absolute value of the number from -2 should be 4

$|x--2|=4$

$|x+2|=4$

So either

$(x+2)=4$

$x=4-2=2$

$-(x+2)=4$

$x+2=-4$

$x=-4-2$

$x=-6$

[tex] The numbers are 2 and -6 [/tex[

They are four units away from -2 on the number line

4 0

3 years ago

2x – 5x + 16 = –3(x – 6)

Troyanec [42]

Answer:

0=2

There are no solutions

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

sam's music teacher told him to practice his trombone for X amount of hours. he spent X amount of hours practicing did he practi

Crank

Yes if he practices X amount of hours and is told to practice X amount of hours than he practiced long enough

4 0

3 years ago

Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis

lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0

2 years ago